Question: $x^3y-2x^2+y^4=8$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{4x}{3x^2+4y^3}$ (Choice B) B $\dfrac{x^3+4y^3}{4x-3x^2y}$ (Choice C) C $\dfrac{4x}{3x^2+4y^3}$ (Choice D) D $\dfrac{4x-3x^2y}{x^3+4y^3}$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $x^3y-2x^2+y^4=8$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} x^3y-2x^2+y^4&=8 \\\\ \dfrac{d}{dx}(x^3y-2x^2+y^4)&=\dfrac{d}{dx}(8) \\\\ \dfrac{d}{dx}(x^3y)-\dfrac{d}{dx}(2x^2)+\dfrac{d}{dx}(y^4)&=0 \\\\ 3x^2\cdot y+x^3\cdot\dfrac{dy}{dx}-4x+4y^3\cdot\dfrac{dy}{dx}&=0 \\\\ 3x^2y+x^3\cdot\dfrac{dy}{dx}-4x+4y^3\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 3x^2y+x^3\cdot\dfrac{dy}{dx}-4x+4y^3\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(x^3+4y^3)&=4x-3x^2y \\\\ \dfrac{dy}{dx}&=\dfrac{4x-3x^2y}{x^3+4y^3} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{4x-3x^2y}{x^3+4y^3}$.